Sunday, 18 September 2016

limit with summation and product





Given L=limnnr=1(1r!ri=1(i2+13)). then L is




Try: limnnr=11r![(12+13)(22+13)(r2+13)]



could some help me to solve it, thanks


Answer



For any |a|<1,n=11n!nk=1(ak+b)=n=1(a)nn!nk=1(bak)=n=1(a)nn!(ba1n).


Note that |a|<1, by the generalized binomial theorem,(1a)ba1=n=0(a)nn!(ba1n),

thusn=11n!nk=1(ak+b)=n=0(a)nn!(ba1n)1=(1a)ba11.



In this question, take a=12 and b=13, thenL=n=11n!nk=1(k2+13)=2531,


and [L]=2.


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